Is it possible to create an edge in coin toss bet?

In the above 2 links, it's mentioned that if we toss a fair coin ten times then the exact probability that NO consecutive heads come up in succession (i.e. n = 10 and k = 2) is
p(10,2) = 0.14...

So, the probability of at least one pair of heads, or tails, in 10 tosses is approx 1-0.14 ~0.86

Can we somehow create a strategy with this probability to have a positive expectation game?

Event A: prob. of next toss is always 0.5
Event B: prob. of having consecutive heads or tails in cluster of 10 tosses is > 0.86

Considering event B only as our universe, we employ a cancellation scheme such that in a cluster/interval of 10 tosses, when we first encounter a win of 1unit, on the next toss, we bet Sum of total losses(within that cluster).
If this bet win, we have a net win of 1unit & we restart count of cluster of 10 with first 1unit bet again. If not, we bet 1unit until we encounter another win and ,on the next toss, bet sum of total losses(within that cluster) again.
This betting pattern will remain within 10 tosses, then we restart/count all over again regardless its a net win or loss within 10 tosses.

With event B .i.e. Group of 10 tosses as an "event" with prob>0.86 rather than event A i.e. single toss as an "event", I wonder if we can achieve a positive expectation with some kind of edge. If so, then any trading system with win/loss hit ratio of 50% can be profitable.

Since "prob. of next toss is always 0.5," it would not matter when you place your bet. It would not matter if it is after a win or a loss.
As for betting the sum of total losses, this is similiar to a "martingale" betting scheme, in that the bet would be increased after a loss.

So i guess considering the single probability (in groups of 10) as a solo event is also a fallacy? Law of large number discounts "Everything" else.
This is Not similar to martingale in that the key is Not about increasing bet size after a win Or loss but rather on the condition that "If a coin was tossed 10 x 10 times, i.e. Ten events B, there would be at least 8 out of 10 events, a consecutive head appearing within an event"...
.i.e. A 86% prob. single event